Holographyforum.org / holowiki.org

[quote=BobH post_id=73354 time=1735157334 user_id=2166]
[quote=Din post_id=73352 time=1735153245 user_id=2357]

So, let's see, ... the Center of All Knowledge.
[/quote]

Just more insignificant word salad from the resident troll. :roll:

The benefit I get from these illogical and childish responses to my comments here is that sometimes they incite me to look up fundamental things and that can lead to improvements in the holograms I'm currently making, or solutions to problems I currently face. ;)
[/quote]

So, you have a reference that a scalar equation can have a vector-based value? Any reference at all? I'd say it's pretty fundamental to deny all the world's greatest optical physicists and engineers who all say:

1. Holography is based on wavefront reconstruction. You, apparently, think that holography has nothing to do with wavefront reconstruction!
https://pubs.aip.org/physicstoday/article-abstract/18/8/26/424838/Wavefront-reconstruction-photographyNew-interest?redirectedFrom=fulltext "New interest has arisen in the wavefront reconstruction process of Gabor." NO! NO! According to the Font of All Knowledge, wavefront reconstruction is not holography. Gabor didn't invent holography - Bob Hess did!!!
2. The diffraction efficiency of a thick hologram is based on index modulation, material thickness and recording wavelength and slant factors, [i]not[/i] polarisation. This is the basis of [i]all[/i] calculations by the top optical engineers and physicists on the planet.

To repeat, the DE of a hologram (Kogelnik eq 57) is

η = 1/{1 + 1/sh²√(ν²)}

where ν = iπ(δn)d/{(λ√c(r)c(s)}
I repeat, which variable represents polarisation? The world awaits your stunning discovery!

Oh, by the way, can you quote a reference to your odd hypothesis that DE is proportional to polarisation? From peer-reviewed journal? Not your friend that thinks the world is flat.

[quote=Din post_id=73352 time=1735153245 user_id=2357]

So, let's see, ... the Center of All Knowledge.
[/quote]

Just more insignificant word salad from the resident troll. :roll:

The benefit I get from these illogical and childish responses to my comments here is that sometimes they incite me to look up fundamental things and that can lead to improvements in the holograms I'm currently making, or solutions to problems I currently face. ;)

[quote=BobH post_id=73351 time=1735141248 user_id=2166]
[quote=Din post_id=73350 time=1735067956 user_id=2357]
In re-reading this explanation of diffraction efficiency, I overlooked the obvious, .... Details here: https://onlinelibrary.wiley.com/doi/abs/10.1002/j.1538-7305.1969.tb01198.x
[/quote]

Just more obfuscation to distract from your misunderstanding of the topic of this thread, the nature of Fresnel reflections, and the dependency of diffraction efficiency on polarization of the reconstruction light in thick volume materials like covestro's photopolymer.
[/quote]

So, let's see, Kogelnik's expression for DE of a reflection hologram with dephasing parameter ξ²(eqn 57):

η = 1/{1 + (1 + ξ²/ν²)/sh²√(ν² - ξ²)}

If the hologram is reconstructed on-Bragg, the dephasing parameter vanished and ξ = 0. Now the Kogelnik expression for the diffraction efficiency is:

η = 1/{1 + 1/sh²√(ν²)}

Now, I have never been able to determine the factor representing polarisation, which would, of course be a vector. There are no vector variables in this equation. Let's say that η is polarisation dependant, I assume you mean [i]linear[/i] polarisation (For Bob's sake, let me define 'linear' as meaning a straight line. You know Bob, where there is no deviation from the straight and narrow - sorry Bob, 'deviation' is a three syllable word; never mind). So, one assumes that there is a η(polarisation). But, there are an infinite number of linear polarisations, pick any angle from 0 to 2π - there are an infinite number of them!. So, any given hologram has an infinite number of efficiencies? You know, you should really write this up and publish in "Optics" or "Nature". The world awaits!

Oh! forgot, in the Bob Hess universe ALL physics is '/goobledygook', ALL mathematics is 'goobledygook'. The ultimate in science denial.

But wait! Lawrence Livermore Labs uses holography ( https://www.llnl.gov/article/48521/holographic-plasma-lenses-ultra-high-power-lasers ), and, I've talked to some of their physicists, and they do model their holograms on Kogelnik. But, in the Bob Hess Universe, it's all 'goobledygook" and Kogelnik is all wrong. In the Bob Hess Universe, Bob Hess has Supreme Knowledge and everyone else employing Kogelnik is simply obfuscation ( a 4 syllable word! I'm impressed!).

NASA uses holography (and I've worked on some of them) https://techport.nasa.gov/projects/5664. I know they use the Kogelnik equations because we at POC sent them a report where I did the theoretical work on the holograms using Kogelnik.But, in the Bob Hess Universe, it's all 'goobledygook" and Kogelnik is all wrong. In the Bob Hess Universe, Bob Hess has Supreme Knowledge and everyone else employing Kogelnik is simply obfuscation ( a 4 syllable word! I'm impressed!).

The Department Of Defense uses holography, at POC they were one of our clients. I know they use the Kogelnik equations because we at POC sent them a report where I did the theoretical work on the holograms using Kogelnik.But, in the Bob Hess Universe, it's all 'goobledygook" and Kogelnik is all wrong. In the Bob Hess Universe, Bob Hess has Supreme Knowledge and everyone else employing Kogelnik is simply obfuscation.

Oh We could go on. National labs the world over, military applications the world over, medical applications etc etc. But, in the Bob Hess Universe, it's all 'goobledygook" and Kogelnik is all wrong. In the Bob Hess Universe, Bob Hess has Supreme Knowledge and everyone else employing Kogelnik is simply obfuscation

So, do you have a reference for the polarisation dependance of DE in any peer-reviewed technical journal or text book? Where are these 'many books'?

I forgot Bob Hess is the Font of All Knowledge and cannot be questioned. Despite having NO valid references and NO mathematical or physical argument.

Oh, just mentioning, the earth [b]is[/b] round,and we [b]did[/b] go to the moon. The Covid vaccine [b]does not [/b] autism. Almost forgot, the earth actually travels around the sun. The earth is not the center of the universe. It's been shown mathematically. But, in the Bob Hess universe All mathematics is 'goobledygook' and the earth is actually the centre of the universe (because Bob Hess lives there, of course!)

EDIT: Merry Christmas to the Font of All Wisdom and the Center of All Knowledge.

[quote=Din post_id=73350 time=1735067956 user_id=2357]
In re-reading this explanation of diffraction efficiency, I overlooked the obvious, .... Details here: https://onlinelibrary.wiley.com/doi/abs/10.1002/j.1538-7305.1969.tb01198.x
[/quote]

Just more obfuscation to distract from your misunderstanding of the topic of this thread, the nature of Fresnel reflections, and the dependency of diffraction efficiency on polarization of the reconstruction light in thick volume materials like covestro's photopolymer.

In re-reading this explanation of diffraction efficiency, I overlooked the obvious, viz the difference between the [i]recording of a hologram[/i] and the [i]reconstruction of it[/i] I was pretty sick last week, so the DE of my brain function was compromised. Yesterday morning, Joy saw me picking up where I left off reading "The Birth of Classical Europe. A history from Troy to Augustine" and quipped, "Brain's back online I see".

At any rate, when you're recording a hologram, the beam ratio is important, as is polarisation because you're [i]creating a diffractive structure[/i]. You're creating a structure - the diffractive structure - based on the modulation, the variation between the actinic to non-actinic reaction. Polarisation is important in these circumstances because the modulation is proportional to the cosine of the polarisation angles of the two beams. If, for example, the polarisation of the (two) beams were perpendicular, then, since cos(90) = 0, there'd be no modulation. Also, if either beam is elliptically polarised there'd be no modulation. But, the important point here is that you're creating a diffractive structure, there is no diffractive structure until the process is complete. Even then, modulation is very low if further processing is required. So, for example, the modulation for unprocessed silver and DCG holograms is very, very low and is known as the 'latent image'. In silver halide, for example, development amplifies the modulation by some 10,000,000 times. All of this is not quite true for ferro-electric crystals like lithium niobate, In which case, the actinic reaction is one of electron mobilisation - the energy from the light causes electron movement, which in turn causes a variation in charge density, and, the refractive index is proportional the charge density, or the modern photopolymers, where the light causes monomers to polymerise, and density difference between the polymers and the monomers cause an index modulation; details on the Bayer photopolymer here: https://pmc.ncbi.nlm.nih.gov/articles/PMC6418958/. But, let me again stress, these parameters determine the modulation of the diffractive (Bragg) structures.

However, this is very different to reconstructing an already recorded hologram. In this case, the modulation is already set in stone, as it were. Now, we have a finished diffractive (Bragg) structure. As explained, once light enters such a structure, there is an interchange of energy between the entering light and the (already finished) diffractive structre. This energy interchange "creates" new light fields - the diffracted light field - with new properties of intensity and direction. The only parameters in this interchange is the energy of the light and the modulation of Bragg structure. The efficiency of the energy interchange - the 'diffractive efficiency' - is based on how much energy goes to the new light field and is [i]defined[/i] as the ratio of the energy of the original lightfield and the new lightfield.

So, to recap, while [i]recording a hologram[/i], you're creating a modulation. But, you cannot determine the efficiency of that modulation while you're creating it; you cannot lift yourself by tugging at your boots. [i]Once created,[/i] the modulation is built-in. Now, the efficiency of that modulation is determined by an energy interchange between the structure and the incoming (scalar) light field. Details here: https://onlinelibrary.wiley.com/doi/abs/10.1002/j.1538-7305.1969.tb01198.x

[quote=BobH post_id=73348 time=1734819596 user_id=2166]
"Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate. Therefore, Fresnel reflection need to be considered, since such reflection diminish the actual radiation entering the plate. So, for example, if you were recording at angles of 0 and 80 degrees, then very little of the beam power measured at 80 degrees is actually entering the plate. The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate."

Those were your words in the thread of Sun Jan 23, 2022 7:50 am. Something went wrong with my quoting it from there.

You're now actually arguing with yourself.
[/quote]
Oh my Gawd!
The meter reading has nothing to do with the mechanism of the DE. Whatever method light is prevented from entering the material, only light that has entered the material counts toward the DE of the hologram.If light does not go into the medium, then any light that doesn't go into the medium is irrelevant. In this case, any light that diffracts out of the medium is what the DE is based on, DE, η = |S(d)|²/|R(0)|² (for a transmission hologram). The action of the complex function S(d) - light leaving the medium - as a ratio of light eentering the medium, the complex field R(0) gives the DE. The intensity of the light exiting the medium, as a ratio of the light entering the medium.

Thus, if light is prevented from entering the medium the DE of the medium is superfluous to the light that has not entered and R(0) = 0 for superfluous light.

[quote=BobH post_id=73348 time=1734819596 user_id=2166]
The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate.
[/quote].
It could be dust in the room, it doesn't matter; if light does not enter the medium, it doesn't count as part of the DE. This is what the DE is.

I can see how you might be confused (happens a lot, doesn't it). I'll try and keep it really, really simple. !0 units of light approach the material. Through various mechanisms, 4 units of light do not enter the medium - it could be polarisation, it could be scattering (oops! Sorry! I said I'd keep it simple) So, if you lose 4 units of light through various mechanisms (aliens, scattering, The Great Kogelnik in the sky - whatever). Now, 6 units of light enter the medium. You can measure this with a meter, even a meter with a polartiser on the head. Now, [i]provided you keep the polariser in place[/i] you measure the light exiting the medium. will be about 6 units. Now, if the light leaving the diffracted light in the direction is measured at 3 units, the hologram is 3/6 = 0.5, and your hologram is 50% efficient.

"Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate. Therefore, Fresnel reflection need to be considered, since such reflection diminish the actual radiation entering the plate. So, for example, if you were recording at angles of 0 and 80 degrees, then very little of the beam power measured at 80 degrees is actually entering the plate. The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate."

Those were your words in the thread of Sun Jan 23, 2022 7:50 am. Something went wrong with my quoting it from there.

You're now actually arguing with yourself.

[quote=BobH post_id=73341 time=1734794655 user_id=2166]
Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate.
[/quote]
No true, the fringe structure exists in space whether or not there's a plate. At any given point in space, in one dimension, the fringe structure is given by the visibility function:
I = I(1) + I(2) + 2√(I(1)I(2)) cos (φ), where I(1) and I(2) are beam intensities and φ is the phase difference .
The Fringe Visibity function is given by:
V = {I(1)(max) - I(min)}/I(max) + I(min). (https://phys.libretexts.org/Bookshelves/Optics/BSc_Optics_(Konijnenberg_Adam_and_Urbach)/05%3A_Interference_and_coherence/5.10%3A_Fringe_Visibility)
These functions have nothing to do with any plate, you can shine these light on a wall, independent of any 'plate'. I'm not sure what you're getting at, since I don'tunderstand "wooly thinking"

[quote=BobH post_id=73341 time=1734794655 user_id=2166]
since such reflection diminish the actual radiation entering the plate.
[/quote]
Exactly! That's why efficiency is not polarisation dependent. The only parameters are the light going in and the light going out. Any light that does not go into the material, whether because of the Fresnel equations, or just dust in the room, does not take part in the efficiency calculation.

[quote=BobH post_id=73341 time=1734794655 user_id=2166]

Pure evidence that you'll say whatever you think is necessary to troll and argue with anyone who challenges your generally poor advice (despite your correct point about the real value of the beam ratio above).
[/quote]
Ad hominem attack, which I'm reporting. By the way, all this "evidence" is purely hand-waving. Do you a mathematical argument?

[quote=Din post_id=73329 time=1734628346 user_id=2357]

However, there is no such thing as "Fresnel reflection", as opposed to any other kind of reflection; the phenomena of reflection is defined by the equation:i = r ( angle of incidence = angle of reflection ).
[/quote]

"No such thing as "Fresnel reflection"?!? You didn't think so almost three years ago when you said the following here:

"BobH wrote: ↑Sun Jan 23, 2022 7:50 am
"Beam ratio is independent of the orientation of a recording plate, as it determines the fringe visibility in space."

Not true. The beam ratio is determined by the actual actinic radiation actually entering the plate. Therefore, Fresnel reflection need to be considered, since such reflection diminish the actual radiation entering the plate. So, for example, if you were recording at angles of 0 and 80 degrees, then very little of the beam power measured at 80 degrees is actually entering the plate. The meter reading itself is independent of polarisation and therefore cannot determine how much power is entering the plate."

Pure evidence that you'll say whatever you think is necessary to troll and argue with anyone who challenges your generally poor advice (despite your correct point about the real value of the beam ratio above).

Din,

And they say alcohol is bad for you....

BobH,

Mode hopping is it. The He-Ne is crystal clear.

Cheers.